A bounded complex linear functional on a subspace of a complex normed space extends with the same norm

Let $u:=\mathrm{Re}{f}_0$. By [L2], $u$ is real linear on the underlying real subspace $M$. Also $\mathrm{\mid }u(m)\mathrm{\mid }\le \mathrm{\mid }{f}_0(m)\mathrm{\mid }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }m\mathrm{\parallel }(m\in M),$ so $u$ is bounded with $\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }$.

Apply [L1] to the underlying real normed spaces. This yields a bounded real linear functional $U:X\to \mathbb{R}$ extending $u$ and satisfying $\mathrm{\parallel }U\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }$. Define $F(x):=U(x)-iU(ix)(x\in X)\mathrm{.}$ By [L2], $F$ is complex linear and $\mathrm{Re}F=U$.

For $m\in M$, the equality $U{\mathrm{\mid }}_M=u$ and [L2] give $F(m)=u(m)-iu(im)=f_0(m)\mathrm{.}$ So $F$ extends $f_0$.

Fix $x\in X$. Choose $\theta \in \mathbb{R}$ so that $e^{i\theta }F(x)=\mathrm{\mid }F(x)\mathrm{\mid }$ is a nonnegative real number. Since $\mathrm{Re}F=U$ and $F$ is complex linear, $\mathrm{\mid }F(x)\mathrm{\mid }=\mathrm{Re}(e^{i\theta }F(x))=\mathrm{Re}F(e^{i\theta }x)=U(e^{i\theta }x)\mathrm{.}$ Therefore $\mathrm{\mid }F(x)\mathrm{\mid }\le \mathrm{\parallel }U\mathrm{\parallel }\mathrm{\parallel }{e}^{i\theta }x\mathrm{\parallel }=\mathrm{\parallel }U\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel },$ where the last inequality uses step 1.1 and $\mathrm{\parallel }U\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }$. Hence $\mathrm{\parallel }F\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }$.

Since $F{\mathrm{\mid }}_M=f_0$, every $m\in M$ with $\mathrm{\parallel }m\mathrm{\parallel }\le 1$ satisfies $\mathrm{\mid }{f}_0(m)\mathrm{\mid }=\mathrm{\mid }F(m)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }$. Taking the supremum over the unit ball of $M$ gives $\mathrm{\parallel }{f}_0\mathrm{\parallel }\le \mathrm{\parallel }F\mathrm{\parallel }$. Combined with step 3.2, this yields $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }$.